Question

At 60oC the value of Kw is 90.5 x 10-4. Considering this, what is the calculated value for the pH of solution made by dissolving 1.00g of sodium hydroxide in enough water to make 500ml of solution at this temperature?

Answer 1

There is an error in the formulation of the problem. Kw cannot be equal to 10^-4, it must be equal to 10^-14.

n_NaOH = 1(g)/40(g/mol) = 0.025 mol

C_NaOH = n/V = 0.25 mol / 0.5 l = 0.05 mol/l

C_[H+] = K_w / C_NaOH = 90.5•10^-14/0.05 = 1.81•10^-11 mol/l

pH = -lgC_[H+] = 10,74

Answer is pH=10,74